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RC column design to SANS 10100

A short braced 400 mm square column under axial load and bending: slenderness, minimum eccentricity, axial capacity, the N–M interaction by strain compatibility, steel limits and links. Every number below comes from the free STATIX RC column calculator and was checked by hand.

1. The problem

InputValue
Section400 × 400 mm rectangular column, cover 40 mm
Design loads (ultimate, already factored)N = 1,500 kN, Mx = 80 kNm, My = 0, V = 0
Effective lengthle = 3.0 m about both axes, braced
Materialsfcu = 30 MPa, fy = 450 MPa

The calculator asks for the effective length, not the clear height. You find it as le = β × lo, with β from Table 18 of SANS 10100 for a braced column (0.75 to 1.0, depending on how stiff the beams at each end are). Here le = 3.0 m is given.

The N-M interaction diagram of the column with 6Y12 bars, N_uz 2,380 kN at the top and the design point (80 kNm, 1,500 kN) inside the envelope, from the STATIX RC column calculator
The STATIX RC column calculator with these inputs (6Y12, Y10 links at 125 mm): the N–M interaction diagram, Nuz = 2,380 kN, with the design point (80 kNm, 1,500 kN) inside the envelope.

2. Slenderness (SANS 10100 cl. 4.7.1.4)

le/h = 3,000 / 400 = 7.5 le/b = 3,000 / 400 = 7.5 Limit for a braced column: 17 − 7M1/M2 = 17 − 7 × 1 = 10 (M1/M2 = 1, see below) 7.5 ≤ 10 → short column, no additional moment

Check both directions. A column can be short about one axis and slender about the other. This one is square, so the two ratios are the same.

The braced limit depends on the ratio of the smaller to the larger end moment. With the end moments unknown, the calculator takes M1/M2 = 1, which gives the lowest limit, 10. With real end moments in double curvature the limit is higher. At 7.5 this column is short either way.

3. Minimum eccentricity (cl. 4.7.2.3)

No column is designed for pure axial load. The design moment must be at least N × emin, about each axis separately.

emin = 0.05h, but not more than 20 mm = min(0.05 × 400, 20) = 20 mm N · emin = 1,500 × 0.020 = 30 kNm About x: 30 kNm < 80 kNm → design moment M = 80 kNm About y: My = 0, so the section must still carry 30 kNm about y on its own (step 6)

4. Steel: minimum governs

Ac = 400 × 400 = 160,000 mm² Asc,min = 0.4% × 160,000 = 640 mm² (Table 23) Try 6Y12: Asc = 6 × 113.1 = 679 mm² = 0.42% (limits 0.4% to 6%) OK

The calculator starts at the minimum steel and adds more only if the column fails. Here the minimum is already enough. It lays the six bars out symmetrically: one at each corner and one in the middle of each face at right angles to the bending, so 3 bars (339 mm²) on the top face and 3 on the bottom face.

The calculator measures the bar depth as cover + 10 mm link + 8 mm (half a 16 mm bar) = 58 mm, even for 12 mm bars. The true depth is 56 mm, so it slightly under-rates the section. That is on the safe side.

5. Axial capacity Nuz (cl. 4.7.3.1)

Nuz = 0.45 fcu (Ac − Asc) + 0.75 fy Asc (eq. 11, net concrete area) = 0.45 × 30 × (160,000 − 679) + 0.75 × 450 × 679 = 2,151 + 229 = 2,380 kN N = 1,500 kN ≤ 2,380 kN OK

SANS 10100 gives Nuz with 0.75fy on the steel; the allowance for the material factor is built in. This is the axial capacity with no moment at all and is the top of the interaction diagram. The real check is the interaction in the next step.

6. N–M interaction by strain compatibility (cl. 4.7.4.1)

SANS 10100 designs a column section with the same assumptions as a beam (cl. 4.3.3.1). Pick a neutral axis depth x. The concrete strain at the compressed face is 0.0035. The steel in compression works up to fyc = fy / (1.15 + fy/2000) = 450 / 1.375 = 327 MPa (SANS 10100-1 figure 2), not 0.87 fy. The stress block is 0.45 fcu (0.447 fcu exactly, 0.67/1.5) over a depth 0.9x. The steel stress follows its strain, up to 0.87 fy = 391.5 MPa. Move x until the internal forces add up to N = 1,500 kN. The moment of those forces about the centre is the moment capacity.

Try x = 295.5 mm: block depth a = 0.9 × 295.5 = 265.9 mm Concrete Cc = 0.447 × 30 × 400 × 265.9 = 1,426.4 kN Top bars d = 58 mm: ε = 0.0035 × (295.5 − 58)/295.5 = 0.0028 > fyc/Es → 327.3 MPa × 339 = +111.0 kN Bottom bars d = 342 mm: ε = −0.00055 → −110.1 MPa × 339 = −37.4 kN ΣN = 1,426.4 + 111.0 − 37.4 = 1,500 kN (matches N) MRd = 1,426.4 × (200 − 265.9/2) + 111.0 × (200 − 58) + 37.4 × (342 − 200) = 95.6 + 15.8 + 5.3 = 116.7 kNm Utilisation = 80 / 116.7 = 0.69 OK

At 1,500 kN the neutral axis sits deep in the section and the bottom steel is barely in tension. That is normal for a column carrying a lot of axial load. The biggest moment this section can carry is about 142 kNm, at an axial load near 1,035 kN.

About y (minimum eccentricity only): the same 6 bars now lie as 2 on each face and 2 on the centre line MRd,y at 1,500 kN = 111.5 kNm ≥ 30 kNm OK (27%)

For bending about y the two middle bars sit on the centre line, where they carry axial load but add no moment, so the capacity about y is lower than about x. Bending can reverse in a real frame; the bars are the same on opposite faces, so the capacity is the same either way round.

7. Links (cl. 4.11.4.5.1)

Link size ≥ max(6 mm, φbar/4) = max(6, 3) = 6 mm → Y10 provided OK Spacing ≤ 12 × φbar = 12 × 12 = 144 mm Provide Y10 at 125 mm (144 rounded down to a 25 mm step), one perimeter link

The 12φ rule stops the compression bars from buckling between links. The code also asks that every corner bar and every alternate bar is held by a link corner of 135° or less, and that every other bar is within 150 mm of a held bar. One perimeter link holds the 4 corner bars. The bar in the middle of each face is not held, but both its neighbours are, and it is 142 mm centre to centre from them: 130 mm clear, inside 150 mm. So one perimeter link is enough.

8. Shear (cl. 4.7.4.5)

V = 0 → no shear demand, links from step 7 only vmax = min(0.75√30, 4.75) = 4.11 MPa

If there were shear, it would be checked as for a beam, with vc from eq. (2) raised for the axial load by 0.6NVh/(AcM) (cl. 4.3.4.4, eq. 2(a)), with Vh/M not more than 1.

9. Result

CheckResult
Slendernessle/h = 7.5 ≤ 10: short
Axial capacity1,500 ≤ Nuz = 2,380 kN
N–M interaction about x80 ≤ 116.7 kNm at 1,500 kN (69%)
Minimum eccentricity about y30 ≤ 111.5 kNm (27%)
Longitudinal steel6Y12 = 679 mm² = 0.42% (0.4% to 6%)
LinksY10 at 125 mm, one perimeter link
Quantities (3 m length)6Y12 × 3.0 m and 25 links of 1,440 mm: about 38 kg of steel

The bar lengths in the schedule are the effective length only. Add laps or starter bars for the real storey height.

Run this example yourselfThe free STATIX RC column calculator opens with exactly these inputs. Change any value and the interaction diagram and every step update, with a printable calc sheet.
Open the free RC column calculator

A worked example for learning, not a design. Check every value against the current code and your own project, and have the design checked and signed by a registered professional engineer.

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